kesempatan kali ini saya akan memaparkan contoh perhitungan konstruksi portal dengan metode takabeya.
Dik : seperti tergambar
| KEKAKUAN (K) |
NILAI τ | |||
| KA1= | 0,6 | τ1= | -1,313 | |
| K12= | 0,75 | τ2= | 3,500 | |
| K2F= | 0,6 | τ3= | -2,208 | |
| KB3= | 0,64 | τ4= | -4,042 | |
| K13= | 1 | |||
| K34= | 0,8 | NILAI ρ | ||
| K24= | 1 | ρ1= | 4,7 | |
| K4E= | 0,64 | ρ2= | 4,7 | |
| K3C= | 0,96 | ρ3= | 6,8 | |
| K4D= | 0,96 | ρ4= | 6,8 | |
| MOMEN
PRIMER (Ṁ) |
NILAI γ | |||
| ṀA1= | -4,688 | γ12= | 0,160 | |
| Ṁ1A= | 4,688 | γ21= | 0,160 | |
| Ṁ12= | -6,000 | γ13= | 0,213 | |
| Ṁ21= | 6,000 | γ31= | 0,147 | |
| Ṁ2F= | -2,500 | γ24= | 0,213 | |
| ṀF2= | 2,500 | γ42= | 0,147 | |
| ṀB3= | -3,125 | γ34= | 0,118 | |
| Ṁ3B= | 3,125 | γ43= | 0,118 | |
| Ṁ13= | 0 | |||
| Ṁ31= | 0 | MOMEN PARSIL | ||
| Ṁ34= | -5,333 | m1(⁰)= | 0,279 | |
| Ṁ43= | 5,333 | m2(⁰) = | -0,745 | |
| Ṁ24= | 0 | m3(⁰)= | 0,325 | |
| Ṁ42= | 0 | m4(⁰)= | 0,594 | |
| Ṁ4E= | -9,375 | |||
| ṀE4= | 9,375 | |||
| Ṁ3C= | 0 | |||
| ṀC3= | 0 | |||
| Ṁ4D= | 0 | |||
| ṀD4= | 0 | |||
Pemberesan Momen Parsil
| momen Akhir |
|
| MA1= | -4,451 | tm |
| M1A= | 5,160 | tm |
| M12= | -6,129 | tm |
| M21= | 4,856 | tm |
| M2F= | -3,651 | tm |
| MF2= | 1,924 | tm |
| M13= | 0,970 | tm |
| M31= | 0,759 | tm |
| MB3= | -3,008 | tm |
| M3B= | 3,359 | tm |
| M24= | -1,205 | tm |
| M42= | 0,468 | tm |
| M34= | -4,470 | tm |
| M43= | 6,622 | tm |
| M4E= | -8,461 | tm |
| ME4= | 9,832 | tm |
| M3C= | 0,351 | tm |
| MC3= | 0,176 | tm |
| M4D= | 1,371 | tm |
untuk lebih jelas mengenai cara dan rumus-rumus yang dipakai, silahkan di download file di bawah ini
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note :
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